Question

Find p[b] in each case: (a) events a and b are a partition and p[a] = 3p[b]. (b) for events a and b, p[a ∪ b] = p[a] and p[a ∩ b] = 0. (c) for events a and b, p[a ∪ b] = p[a]− p[b].

Answer 1

(a)
`p[b]=(1)/(3)* p[a]`.

(b) p[b]=0

(c)
`p[b]=(1)/(2)* p[a\cap b]`

Explanation:

(a)

`p[a]=3p[b]`

Divide both sides by 3.

`(1)/(3)* p[a]=p[b]`

Therefore
`p[b]=(1)/(3)* p[a]`.

(b)

`p[a\cup b]=p[a]`

`p[a]+p[b]-p[a\cap b]=p[a]`
`[\because P(A\cup B)=P(A)+P(B)-P(A\cap B)]`

`p[a]+p[b]-0=p[a]`
`[\because p[a\cap b]=0]`

`p[b]=p[a]-p[a]`

`p[b]=0`

Therefore
`p[b]=0`.

(c)

`p[a\cup b]=p[a]-p[b]`

`p[a]+p[b]-p[a\cap b]=p[a]-p[b]`
`[\because P(A\cup B)=P(A)+P(B)-P(A\cap B)]`

`p[b]-p[a\cap b]+p[b]=0`

`2p[b]=p[a\cap b]`

`p[b]=(1)/(2)* p[a\cap b]`

Therefore
`p[b]=(1)/(2)* p[a\cap b]`.

Answer 2

Case(a):
`p[b]=(1)/(3)p[a]`

Case(b):
`p[b]=0`

Case(c):
`p[b]=(1)/(2)p[a\cap b]`

Explanation:

Given

(a) events a and b are a partition and p[a] = 3p[b].

(b) for events a and b, p[a ∪ b] = p[a] and p[a ∩ b] = 0.

(c) for events a and b, p[a ∪ b] = p[a]− p[b].

we have to find the p[b] in each case:

Case (a): events a and b are a partition and p[a] = 3p[b].

gives
`p[b]=(1)/(3)p[a]`

Case (b): for events a and b, p[a ∪ b] = p[a] and p[a ∩ b] = 0.

`p[a\cup b]=p[a]` ⇒
`p[a]+p[b]-p[a\cap b]=p(a)` ⇒
`p[b]=0` ∵ p[a ∩ b] = 0.

Case(3): for events a and b, p[a ∪ b] = p[a]− p[b].

p[a ∪ b] = p[a]− p[b]

⇒
`p[a]+p[b]-p[a\cap b]=p[a]-p[b]`

⇒
`2p[b]=p[a\cap b]`

⇒
`p[b]=(1)/(2)p[a\cap b]`